Velocity of a ladybug relative to the ground

AI Thread Summary
The velocity of the ladybug relative to the ground is calculated by combining its velocity with that of the chair. The y-component of the chair's velocity is 25.7 mm/s, while the x-component totals 40.6 mm/s after adding the ladybug's velocity. Using the Pythagorean theorem, the resultant velocity is found to be 48.1 mm/s. The direction is determined to be 32° North of West. The final answer is expressed as 48 mm/s [32° N of W].
TheronSimon
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Homework Statement


A ladybug with a velocity of 10 mm/s [W] crawls on a chair that is being pulled with a velocity of 40 mm/s [40 degrees N of W]. What is the velocity of the ladybug relative to the ground?


Homework Equations





The Attempt at a Solution


finding the y-component
40(sin40) = 25.7

x-component requires the addition of two x-axis component vectors
40cos40 + 10 = 40.6

Now a² + b² = c²
25.7² + 40.6² = c²
c = 48.1mm/s
 
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Hi TheronSimon! :smile:
TheronSimon said:
A ladybug with a velocity of 10 mm/s [W] crawls on a chair that is being pulled with a velocity of 40 mm/s [40 degrees N of W]. What is the velocity of the ladybug relative to the ground?

finding the y-component
40(sin40) = 25.7

x-component requires the addition of two x-axis component vectors
40cos40 + 10 = 40.6

Now a² + b² = c²
25.7² + 40.6² = c²
c = 48.1mm/s

looks ok :smile:

now you just need to state the direction of the velocity :wink:

(btw, maximum 2 significant figures in this case :wink:)
 
so for the direction tan -1 (25.7/40.6) = 32'

so the complete answer is 48 mm/s [W32'N] ?
 
yes :smile:

(probably best to call it "32° N of W", same as in the question :wink:)

(btw, you'll find a ° in the Quick Symbols box on the right of the Reply box)
 
alright thank you very much Tim! :)
 
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